If enthaopy of hydrogenation of `C_(6)H_(6)(l)` into `C_(6)H_(12)(l)` is `-205kJ//mol` and resonance energy of `C_(6)H_(6)(l)` is `-152kJ//mol` then enthaopy of hydrogenation of

Assume `DeltaH_("vap")` of `C_(6)H_(6)(l),C_(6)H_(12)(l)` all equal :

Options

A. `-535.5kJ//mol`
B. `-238kJ//mol`
C. `-357kJ//mol`
D. `-119` kJ/mol

Answer

D

Explanation

Theoretical heat of hydrogenation of benzene = (Actual heat of hydrogenation) + (Resonance energy)

=-205 - 152 =- 357
Enthalpy of hydrogenation of ,
`=(-357)/(3) = - 119` kJ/mol